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RMP 24  34
Posted:
Jul 18, 2009 9:27 AM


Chace's transliteration of the RMP were used to solve this class of algebra problem. Ahmes' thinking is clear, right?
24. x + (1/7)x = 19
(8/7)x = 19,
x = 133/8 = 16 + 5/8 = 16 + (4 + 1)/8 = 16 + 1/2 + 1/8
25. x + (1/2)x = 16;
(3/2) x = 16,
x = 32/3 = 10 + 2/3
26. x + (1/4)x = 15
(7/4)x = 15
x = 60/7 = 8 + 4/7 = 8 + 4/7(4/4) = 8 + (16/28) = 8 + (14 + 2)/28 = 8 + 1/2 + 1/14
27. x + (1/5)x = 21
(6/5)x = 21
x = 105/6 = 17 + 1/2
28. (2/3)x  (1/3)y = 10; (2/3)y = 10
two unknowns.
29. a solution method, not a problem
1 +1/4 + 1/10 = 13 1/2
that scholars classify as a diversion.
31. x + (2/3 + 1/2 + 1/7)x = 33
(97/42)x = 33
x = 1386/97 = 14 + (28/97)
x =14 + (2/97)(56/56) + (26/97)(4/4) = 112/5432 + 104/388 =
14 + (97 + 8 + 7)/5432 + (97 + 4 + 2 + 1)/388
32. x + (2/3 + 1/4)x = 2
(23/12)x = 2
x = 24/23 = 1 + 1/23
33. x + (2/3 +1/2 + 1/7)x = 37
(97/42)x = 37
x = 1554/97 = 16 + 2/97
with 2/97 (56/56) = 112/5432 = (97 + 8 + 7)/56432
34. x + (1/2 + 1/4)x = 10
(7/4)x = 10
x= 40/7 = 5 + 5/7 (4/4) = 5 + 20/28 = 5 + (14 + 4 + 2)/28 = 5 + 1/2 + 1/7 + 1/14
Adding back Ahmes' initial and intermediate steps was required to solve these problems.



